POJ 2392 Space Elevator (多重背包+优化)

[align=center]Space Elevator[/align]

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7378		Accepted: 3464

Description
The cows are going to space! They plan to achieve orbit by building a sort of space elevator: a giant tower of blocks. They have K (1 <= K <= 400) different types of blocks with which to build the tower. Each block of type i has
height h_i (1 <= h_i <= 100) and is available in quantity c_i (1 <= c_i <= 10). Due to possible damage caused by cosmic rays, no part of a block of type i can exceed a maximum altitude a_i (1 <= a_i <= 40000).

Help the cows build the tallest space elevator possible by stacking blocks on top of each other according to the rules.
Input
* Line 1: A single integer, K

* Lines 2..K+1: Each line contains three space-separated integers: h_i, a_i, and c_i. Line i+1 describes block type i.
Output
* Line 1: A single integer H, the maximum height of a tower that can be built
Sample Input

3
7 40 3
5 23 8
2 52 6

Sample Output

48

Hint
OUTPUT DETAILS:

From the bottom: 3 blocks of type 2, below 3 of type 1, below 6 of type 3. Stacking 4 blocks of type 2 and 3 of type 1 is not legal, since the top of the last type 1 block would exceed height 40.
Source
USACO 2005 March Gold

题意：

有一头牛想要建一座石塔，他有很多不同类型的石块。

每类石块有三个属性：1. (h): 石块的高度；2. (a): 石块能达到的最大高度；3. (c): 石块的数量

思路：

采用多重背包。

注意点：在做背包前需要对石块能到达的最大高度（a）进行排序，防止石塔到了一定高度后，剩下的一些石块所能达到的最大高度小于这个高度，那剩下的这些石块也就不能往上放了。排序时按升序排序，先放能达到最大高度小的石块。

优化：直接标记能到达的最高点。

/*************************************************************************
> File Name: C.cpp
> Author: BSlin
> Mail:
> Created Time: 2013年10月01日 星期二 19时18分57秒
************************************************************************/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <iterator>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#define MP make_pair
#define INF (1<<30)
#define PI acos(-1.0)
#define esp 1e-8
const int dx[4]={0,0,0,0};
using namespace std;
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#if defined (_WIN32) || defined (__WIN32) || defined (WIN32) || defined (__WIN32__)
#define LL __int64
#define LLS "%" "I" "6" "4" "d"
#else
#define LL long long
#define LLS "%" "l" "l" "d"
#endif

struct node {
int h,a,c;
bool operator < (const node b) const {
return a < b.a;
}
}data[410];

int dp[40010];

int main(int argc, char** argv) {
//read;
int n,ans;
while(scanf("%d",&n) != EOF) {
memset(dp,0,sizeof(dp));
ans = 0;
for(int i=0; i<n; i++) {
scanf("%d%d%d",&data[i].h,&data[i].a,&data[i].c);
data[i].c = min(data[i].c,data[i].a/data[i].h);
}
sort(data,data+n);
for(int i=0; i<n; i++) {
for(int j=0; j<data[i].c; j++) {
for(int k=data[i].a; k>=data[i].h; k--) {  //不会覆盖到以后的值
dp[k] = max(dp[k],dp[k-data[i].h] + data[i].h);
if(ans < dp[k]) ans = dp[k];
}
}
}
printf("%d\n",ans);
}
return 0;
}

优化后

/*************************************************************************
> File Name: CC.cpp
> Author: BSlin
> Mail:
> Created Time: 2013年10月01日 星期二 21时25分11秒
************************************************************************/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <iterator>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#define MP make_pair
#define INF (1<<30)
#define PI acos(-1.0)
#define esp 1e-8
const int dx[4]={0,0,0,0};
using namespace std;
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#if defined (_WIN32) || defined (__WIN32) || defined (WIN32) || defined (__WIN32__)
#define LL __int64
#define LLS "%" "I" "6" "4" "d"
#else
#define LL long long
#define LLS "%" "l" "l" "d"
#endif

struct node {
int h,a,c;
bool operator < (const node b) const {
return a < b.a;
}
}data[410];

bool dp[40010];
int cnt[40010];

int main(int argc, char** argv) {
//read;
int n,ans;
while(scanf("%d",&n) != EOF) {
memset(dp,0,sizeof(dp));
for(int i=0; i<n; i++) {
scanf("%d%d%d",&data[i].h,&data[i].a,&data[i].c);
}
sort(data,data + n);
dp[0] = 1;
ans = 0;
for(int i=0; i<n; i++) {
memset(cnt,0,sizeof(cnt));
for(int j=data[i].h; j<=data[i].a; j++) {
if(!dp[j] && dp[j-data[i].h] && cnt[j-data[i].h] + 1 <= data[i].c) {
dp[j] = true;
cnt[j] = cnt[j-data[i].h] + 1;
if(ans < j) ans = j;
}
}
}
printf("%d\n",ans);
}
return 0;
}